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Proof step 19 from proof inv-inverts (previous step)

# nσ = ∞, nσ[m] = n2 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1), nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf